JEE MAIN - Mathematics (2025 - 23rd January Evening Shift - No. 4)
Let the point A divide the line segment joining the points $\mathrm{P}(-1,-1,2)$ and $\mathrm{Q}(5,5,10)$ internally in the ratio $r: 1(r>0)$. If O is the origin and $(\overrightarrow{\mathrm{OQ}} \cdot \overrightarrow{\mathrm{OA}})-\frac{1}{5}|\overrightarrow{\mathrm{OP}} \times \overrightarrow{\mathrm{OA}}|^2=10$, then the value of r is :
$\sqrt7$
14
7
3
Explanation
$$\begin{aligned}
&\begin{aligned}
& \mathrm{A} \equiv\left(\frac{5 \mathrm{r}-1}{\mathrm{r}+1}, \frac{5 \mathrm{r}-1}{\mathrm{r}+1}, \frac{10 \mathrm{r}+2}{\mathrm{r}+1}\right) \\
& (\overrightarrow{\mathrm{OQ}} \cdot \overrightarrow{\mathrm{OA}})-\frac{|\overrightarrow{\mathrm{OP}} \times \overrightarrow{\mathrm{OA}}|^2}{5}=10 \quad\text{.... (1)}\\
& \overrightarrow{\mathrm{OQ}} \cdot \overrightarrow{\mathrm{OA}}=\frac{10}{\mathrm{r}+1}(15 \mathrm{r}+1) \\
& |\overrightarrow{\mathrm{OP}} \times \overrightarrow{\mathrm{OA}}|^2=\frac{\mathrm{r}^2}{(\mathrm{r}+1)^2}(800)
\end{aligned}\\
&\text { so by equation (1) }\\
&\begin{aligned}
& \frac{10}{r+1}(15 r+1)-\frac{1}{5} \frac{r^2(800)}{(r+1)^2}=10 \\
& 2 r^2-14 r=0 \\
& r=7, r \neq 0
\end{aligned}
\end{aligned}$$
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